\(\int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [54]

Optimal result

Integrand size = 21, antiderivative size = 32 \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d} \]

[Out]

sin(d*x+c)/a/d-1/2*sin(d*x+c)^2/a/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2746} \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d} \]

[In]

Int[Cos[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

Sin[c + d*x]/(a*d) - Sin[c + d*x]^2/(2*a*d)

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}(\int (a-x) \, dx,x,a \sin (c+d x))}{a^3 d} \\ & = \frac {\sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {(-2+\sin (c+d x)) \sin (c+d x)}{2 a d} \]

[In]

Integrate[Cos[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

-1/2*((-2 + Sin[c + d*x])*Sin[c + d*x])/(a*d)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )}{d a}\)	\(25\)
default	\(\frac {-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )}{d a}\)	\(25\)
parallelrisch	\(\frac {\cos \left (2 d x +2 c \right )-1+4 \sin \left (d x +c \right )}{4 d a}\)	\(28\)
risch	\(\frac {\sin \left (d x +c \right )}{a d}+\frac {\cos \left (2 d x +2 c \right )}{4 a d}\)	\(32\)
norman	\(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(105\)

[In]

int(cos(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/2*sin(d*x+c)^2+sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cos \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right )}{2 \, a d} \]

[In]

integrate(cos(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(cos(d*x + c)^2 + 2*sin(d*x + c))/(a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (22) = 44\).

Time = 1.86 (sec) , antiderivative size = 158, normalized size of antiderivative = 4.94 \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {2 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {2 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} + \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((2*tan(c/2 + d*x/2)**3/(a*d*tan(c/2 + d*x/2)**4 + 2*a*d*tan(c/2 + d*x/2)**2 + a*d) - 2*tan(c/2 + d*x
/2)**2/(a*d*tan(c/2 + d*x/2)**4 + 2*a*d*tan(c/2 + d*x/2)**2 + a*d) + 2*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)*
*4 + 2*a*d*tan(c/2 + d*x/2)**2 + a*d), Ne(d, 0)), (x*cos(c)**3/(a*sin(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )}{2 \, a d} \]

[In]

integrate(cos(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(sin(d*x + c)^2 - 2*sin(d*x + c))/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )}{2 \, a d} \]

[In]

integrate(cos(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(sin(d*x + c)^2 - 2*sin(d*x + c))/(a*d)

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sin \left (c+d\,x\right )\,\left (\sin \left (c+d\,x\right )-2\right )}{2\,a\,d} \]

[In]

int(cos(c + d*x)^3/(a + a*sin(c + d*x)),x)

[Out]

-(sin(c + d*x)*(sin(c + d*x) - 2))/(2*a*d)